已知 a+b = c+d,a^3+b^3 = c^3+d^3
a^3+b^3 = (a+b)(a^2-ab+b^2) = (c+d)(c^2-cd+d^2) = c^3+d^3
所以 a^2-ab+b^2 = c^2-cd+d^2 -------(1)
(a+b)^2 = (c+d)^2
所以 a^2+2ab+b^2 = c^2+2cd+d^2 ----(2)
(2)-(1) 3ab = 3cd,所以 ab = cd ----(3)
(3) 代入 (2) a^2+b^2 = c^2+d^2 -----(4)
用归纳法来做:
a^1 + b^1 = c^1 + d^1
a^2 + b^2 = c^2 + d^2
ab = cd
假设 a^n + b^n = c^n + d^n
所以 (a+b) (a^n + b^n) = (c+d)(c^n+d^n)
而 (a+b) (a^n + b^n) = a^(n+1) + ab^n + (a^n)b + b^(n+1)
= a^(n+1) + ab^n + (a^n)b + b^(n+1)
= a^(n+1) + ab[a^(n-1) + b^(n-1)] + b^(n+1)
(c+d)(c^n+d^n) = c^(n+1) + cd[c^(n-1) + d^(n-1)] + d^(n+1)
所以 a^(n+1) + b^(n+1) = c^(n+1) + d^(n+1)
由此类推,可以证明 a^2009 + b^2009 = c^2002 + d^2009